We can use the equation of a curve in polar coordinates to compute
some areas bounded by such curves.
The basic approach is the same as with any application of integration:
find an approximation that approaches the true value. For areas in
rectangular coordinates, we approximated the region using rectangles;
in polar coordinates, we use sectors of circles, as depicted in
figure 10.3.1. Recall that the
area of a sector of a circle is $\ds \alpha r^2/2$, where $\alpha$ is the
angle subtended by the sector. If the curve is given by $r=f(\theta)$,
and the angle subtended by a small sector is $\Delta\theta$,
the area is $\ds (\Delta\theta)(f(\theta))^2/2$.
Thus we approximate the total area as
$$\sum_{i=0}^{n-1} {1\over 2} f(\theta_i)^2\;\Delta\theta.$$
In the limit this becomes
$$\int_a^b {1\over 2} f(\theta)^2\;d\theta.$$

Example 10.3.2 We find the area between the circles $r=2$ and
$r=4\sin\theta$, as shown in figure 10.3.2.
The two curves intersect where $2=4\sin\theta$, or $\sin\theta=1/2$,
so $\theta=\pi/6$ or $5\pi/6$. The area we want is then
$$
{1\over2}\int_{\pi/6}^{5\pi/6}
16\sin^2\theta-4\;d\theta={4\over3}\pi + 2\sqrt{3}.
$$

Figure 10.3.2. An area between curves.

This example makes the process appear more straightforward than it
is. Because points have many different representations in polar
coordinates, it is not always so easy to identify points of
intersection.

Example 10.3.3 We find the shaded area in the first graph of
figure 10.3.3 as the difference
of the other two shaded areas. The cardioid is $r=1+\sin\theta$ and
the circle is $r=3\sin\theta$. We attempt to find points of intersection:
$$\eqalign{
1+\sin\theta&=3\sin\theta\cr
1&=2\sin\theta\cr
1/2&=\sin\theta.\cr}
$$
This has solutions $\theta=\pi/6$ and $5\pi/6$; $\pi/6$ corresponds to
the intersection in the first quadrant that we need. Note that no
solution of this equation corresponds to the intersection point at the
origin, but fortunately that one is obvious. The cardioid goes through
the origin when $\theta=-\pi/2$; the circle goes through the origin at
multiples of $\pi$, starting with $0$.

Now the larger region has area
$$
{1\over2}\int_{-\pi/2}^{\pi/6} (1+\sin\theta)^2\;d\theta=
{\pi\over2}-{9\over16}\sqrt{3}
$$
and the smaller has area
$$
{1\over2}\int_{0}^{\pi/6} (3\sin\theta)^2\;d\theta=
{3\pi\over8} - {9\over16}\sqrt{3}
$$
so the area we seek is $\pi/8$.